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This is my functional series: $$\sum_{k=1}^\infty x^k \tan(\frac{x}{2^k}) $$

Now, to get the convergence radius (not sure if that's the correct word for it in English), I've been taught that I need to take the absolute value of $a_k$ and solve it. So I tried using the ratio test and got this:

$ \lim \limits_{k \to \infty} \frac{|x \tan(\frac{x}{2^{k+1}})|}{|\tan(\frac{x}{2^{k}})|}$

WolframAlpha tells me that the answer to this is $\frac{|x|}{2}$. I have no idea how to get it, could somebody help me, is there some kind of a formula we can use for tangents?

After that we should get that the convergence radius/area is |x| < 2 (because convergence area for ratio test is < 1).

But after that I have been told that I should double-check manually |x| = 2 and |x| > 2.

Wolfram tells me that $ \lim \limits_{k \to \infty} |2^k \tan(\frac{2}{2^{k}})| = 2$ and therefore x=2 diverges. But again I have no idea how to actually get it.

The limit for $ \lim \limits_{k \to \infty} |(-2)^k \tan(\frac{-2}{2^{k}})|$ doesn't exist so that diverges as well.

Showing |x| > 2 is also quite confusing for me.

PadaKatel
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    One simple way is via equivalents: $\tan u \sim_{u\to0} u$ (or, equivalently, the fact that $\frac{\tan u}{u} \xrightarrow[u\to0]{} \tan^\prime(0) =1$). So, since for any fixed $x$ we have $\frac{x}{2^k} \xrightarrow[k\to\infty]{} 0$, you get $\tan \frac{x}{2^k} \sim_{k\to\infty} \frac{x}{2^k}$. – Clement C. Feb 21 '16 at 20:30
  • This is what I got for the second question: $ \lim \limits_{k \to \infty} |2^k \tan(\frac{2}{2^{k}})| = \lim \limits_{k \to \infty} \frac{\tan(\frac{2}{2^{k}})}{\frac{1}{2^k}} = \lim \limits_{k \to \infty} \frac{\tan(\frac{2}{2^{k}})}{\frac{2}{2^k} 2} = \lim \limits_{k \to \infty}\frac{\tan(\frac{2}{2^{k}})}{\frac{2}{2^k} } \frac{1}{2} = \frac{1}{2}$

    Did I do something wrong? And I still don't quite understand the first one.

     – PadaKatel Feb 21 '16 at 20:40
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    Yes, the denominator is wrong. $\frac{1}{2^k} = \frac{2}{2^k}\cdot \frac{1}{2}$, not $ \frac{2}{2^k}\cdot 2$. I can expand in an answer if you still have some doubts, but the first equation will follow the same approach. – Clement C. Feb 21 '16 at 20:46
  • Ah, thank you. Also, I figured out the first one as well (way too many fractions to write it here at this moment, but in the end I could cross out most things and it simplified to the correct result. Thank you very much.

    But do you have any idea how to show that with x > 2 and x < - 2 diverges?

     – PadaKatel Feb 21 '16 at 20:56
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    The same way will work for $x<-2$, since the ratio test (you have absolute values) will give you a limit $>1$. (For any $x\neq 0$, the limit obtained for the ratio test should be $\left\lvert \frac{x}{2} \right\rvert$. But for $-2$ and $2$, you will get 1, and the ratio test is not enough to conclude. – Clement C. Feb 21 '16 at 21:01
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    But you actually don't need it. By the same technique, you have that for any fixed $x\neq 0$ the general term of the series, $a_k$, is equivalent to $\left(\frac{x}{2}\right)^k$. By comparison, you directly get convergence iff $\lvert x \rvert < 2$ (for $x=\pm 2$, this means $a_k$ does not converge to 0 as $\lvert a_k\rvert \to 1$, so in particualr the series does not converge.) – Clement C. Feb 21 '16 at 21:03
  • I think I get it. I did some tests. If x < 2, then ultimately the limit comes less than 1 to the power of k. Which when k->inf, becomes 0. But when it's over 2 then it is >1 to the power of k and therefore it goes to inf. Thank you. – PadaKatel Feb 21 '16 at 21:16
  • See here. – Mhenni Benghorbal Feb 21 '16 at 22:10

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